给定一个整数N ，任务是构造一个大小为N x N的矩阵M [] [] ，其数字在[1，N ^ 2]范围内，并具有以下条件：
- 矩阵M的元素应为1到N ^ 2之间的整数。
- 矩阵M的所有元素都是成对截然不同的。
- 对于在r到r + a行和c到c + a列(含)中包含单元格的每个正方形子矩阵，对于一些有效整数r，c和a> = 0：M(r，c)+ M(r + a， c + a)是偶数，M(r，c + a)+ M(r + a，c)是偶数。
例子：
Input: N = 2
Output:
1 2
4 3
Explanation:
This matrix has 5 square submatrix and 4 of them ([1], [2], [3], [4]) have a=0 so they satisfy the conditions.
The last square submatrix is the whole matrix M where r=c=a=1. We can see that M(1, 1)+M(2, 2)=1+3=4 and M(1, 2)+M(2, 1)=2+4=6 are both even.
Input: N = 4
Output:
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13
方法：我们知道两个数字的和是偶数，即使它们的奇偶性相同。假设M (i，j)的奇偶校验是奇数，这意味着M (i + 1，j + 1) ，M (i + 1，j-1) ，M (i-1，j + 1) ) ，M (i-1，j-1)必须为奇数。
下面是N = 4以生成大小为4×4的矩阵的说明：
因此，从以上插图中，我们必须在棋盘格图案中填充矩阵。我们可以通过两种方式填充它：
- 所有黑色单元格都有一个奇数整数，所有白色单元格都有一个偶数整数。
- 所有黑色单元格都有一个偶数整数，所有白色单元格都有一个奇数整数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the desired matrix
void UniqueMatrix(int N)
{
int element_value = 1;
int i = 0;
// element_value will start from 1
// and go up to N ^ 2
// i is row number and it starts
// from 0 and go up to N-1
// Iterate ove all [0, N]
while(i < N)
{
// If is even
if(i % 2 == 0)
{
for(int f = element_value;
f < element_value + N; f++)
{
// If row number is even print
// the row in forward order
cout << f << " ";
}
element_value += N;
}
else
{
for(int k = element_value + N - 1;
k > element_value - 1; k--)
{
// If row number is odd print
// the row in reversed order
cout << k << " ";
}
element_value += N;
}
cout << endl;
i = i + 1;
}
}
// Driver Code
int main()
{
// Given matrix size
int N = 4;
// Function call
UniqueMatrix(N);
}
// This code is contributed by chitranayal Java
// Java program for the above approach
public class Gfg
{
// Function to print the desired matrix
public static void UniqueMatrix(int N)
{
int element_value = 1;
int i = 0;
// element_value will start from 1
// and go up to N ^ 2
// i is row number and it starts
// from 0 and go up to N-1
// Iterate ove all [0, N]
while(i < N)
{
// If is even
if(i % 2 == 0)
{
for(int f = element_value;
f < element_value + N; f++)
{
// If row number is even print
// the row in forward order
System.out.print(f+" ");
}
element_value += N;
}
else
{
for(int k = element_value + N - 1;
k > element_value - 1; k--)
{
// If row number is odd print
// the row in reversed order
System.out.print(k+" ");
}
element_value += N;
}
System.out.println();
i = i + 1;
}
}
// Driver Code
public static void main(String []args)
{
// Given matrix size
int N = 4;
// Function call
UniqueMatrix(N);
}
}
// This code is contributed by avanitrachhadiya2155Python3
# Python3 program for the above approach
# Function to print the desired matrix
def UniqueMatrix(N):
element_value = 1
i = 0
# element_value will start from 1
# and go up to N ^ 2
# i is row number and it starts
# from 0 and go up to N-1
# Iterate ove all [0, N]
while(i < N):
# If is even
if(i % 2 == 0):
for f in range(element_value, element_value + N, 1):
# If row number is even print
# the row in forward order
print(f, end =' ')
element_value += N
else:
for k in range(element_value + N-1, element_value-1, -1):
# if row number is odd print
# the row in reversed order
print(k, end =' ')
element_value += N
print()
i = i + 1
# Driver Code
# Given Matrix Size
N = 4
# Function Call
UniqueMatrix(N)C#
// C# program for the above approach
using System;
class GFG
{
static void UniqueMatrix(int N)
{
int element_value = 1;
int i = 0;
// element_value will start from 1
// and go up to N ^ 2
// i is row number and it starts
// from 0 and go up to N-1
// Iterate ove all [0, N]
while(i < N)
{
// If is even
if(i % 2 == 0)
{
for(int f = element_value;
f < element_value + N; f++)
{
// If row number is even print
// the row in forward order
Console.Write(f + " ");
}
element_value += N;
}
else
{
for(int k = element_value + N - 1;
k > element_value - 1; k--)
{
Console.Write(k + " ");
}
element_value += N;
}
Console.WriteLine();
i = i + 1;
}
}
// Driver code
static public void Main ()
{
// Given matrix size
int N = 4;
// Function call
UniqueMatrix(N);
}
}
// This code is contributed by rag2127输出：
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13时间复杂度： O(N ^ 2)
辅助空间： O(N ^ 2)